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LenKa [72]
4 years ago
15

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the c

onstant speed?
Physics
1 answer:
son4ous [18]4 years ago
3 0
Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?
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Suppose your car is equipped with tire pressure monitoring system (TPMS) and the tire pressure is 34.0 psi on a 98F day in calif
natali 33 [55]

Answer:Pressure decreases

Explanation:

Given Temperature decreases form 98 F to 20 F

Kinetic Energy of gas molecules depends on temperature so as the temperature decreases the kinetic energy also decreases and vice-versa.

As gas molecules lose kinetic energy they occupy less volume thereby decreasing the Pressure of tire.  

So the tire pressure monitoring system is illuminated because of the reduction in pressure beyond a defined limit.

5 0
4 years ago
An electron moving with a speed of 8.4 105 m/s in the positive z direction experiences zero magnetic force. When it moves in the
Archy [21]

Answer:

magnitude of the magnetic field must be 1.56 T

Explanation:

As we know that the magnetic force on moving charge is given as

F = q(\vec v \times \vec B)

now when charge is moving along z direction then magnetic force is ZERO

so here we can say that magnetic field must be parallel to velocity of the charge i.e. along z direction

now when charge is moving along x direction then we have

F = qvB

2.1 \times 10^{-13} = (1.6 \times 10^{-19})(8.4 \times 10^5) B

so we have

B = 1.56 T

magnitude of the magnetic field must be 1.56 T

6 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
3 years ago
One of the best ways to avoid injury while riding a bike
mel-nik [20]

know how to ride a bike simply put also wear safety gear like a helmet knee pads things like that

hope this helps

5 0
3 years ago
Read 2 more answers
A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If the separation between the c
Maru [420]

Answer:

The final stored energy will become half.

Explanation:

 We know that stored energy in the capacitor is given as

E=\dfrac{1}{2}CV^2

C=capacitance

V=Voltage difference

E=Energy

C=\dfrac{\varepsilon A}{d}

d=Distance between plates

A=Area

E=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d}\times V^2

If the distance between plates get double ,say d' = 2 d

Then stored energy

E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d'}\times V^2

E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{2d}\times V^2

E'=\dfrac {E}{2}

Therefore the final stored energy will become half.

5 0
4 years ago
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